In (9.3.19), use the dispersion relation to rewrite that integral as an integral with respect to $\omega$.
(a) Interpret the multiplication by $t$ as a derivative with respect to $\omega$ and integrate by parts. Rewrite the result as an integral in $k_3$ to obtain
$$
\beta(\mathbf{y})=\int U_{\mathrm{s}}(\boldsymbol{\xi}, t) B(\mathbf{y}-\xi, t) d \xi_1 d \xi_2 d t .
$$
with
$$
B(\mathbf{y}, t)=\frac{32 y_3}{\pi^2} \int d k_1 d k_2 d k_3 \exp \left(2 i\left\{k_1 y_1+k_2 y_2-k_3 y_3\right\}+i \omega t\right) .
$$
In this form, the reflectivity is seen to be an operator on $U_{\mathrm{s}}$ itself rather than on $t U_{\mathrm{S}}$.
(b) Consider the representation (2.4.6) of $U(\mathbf{x}, t)$, which was shown in (2.4.13) to be the three-dimensional Green's function $G(\mathbf{x}, t)$. Rewrite the denominator of the integrand as $\left(c k \text { sign } k_3\right)^2-\omega^2$, and calculate the integral as a residue sum at the zeros of this expression to obtain
$$
\begin{aligned}
G(\mathbf{x}, t) & =G_{+}(\mathbf{x}, t)+G_{-}(\mathbf{x}, t), \\
G_{ \pm}(\mathbf{x}, t) & = \pm \frac{i c}{16 \pi^3} \int \frac{d k_1 d k_2 d k_3}{k \operatorname{sign} k_3} e^{i\left(k \cdot \mathbf{x} \mp d k t \text { sign } k_3\right)}, \quad t>0 .
\end{aligned}
$$
[Remark: The $k_3$ dependence in the phase is $\left[\left|k_3\right| x_3 \mp c k t\right] \operatorname{sign} k_3$. That is, the Green's function is decomposed by this device into a down-going wave $G_{+}$and an up-going wave $G_{-}$.]
(c) In $G_{+}$, differentiate with respect to $t$ and make the change of variable of integration from $k_3$ to $-k_3$ to obtain the result
$$
\frac{\pi}{c^2} \frac{\partial G_{+}(2 \mathbf{y}, t)}{\partial t}=\frac{2}{y_3} B(\mathbf{y}, t) .
$$